Reatining WAll With SLOPING Bach Fill

7/29/2019 Reatining WAll With SLOPING Bach Fill

1/16

RCC design B.C.Punmia

18.2 TYPE OF RETAINING WALLS

1 Gravity walls

2 Cantilever retaining walls a. T- shaped b. L- shaped

3 Counterfort retainig walls.

4 Buttresssed walls.

The cantilever retaining wall resist the horizontal earth pressure as well as other vertical pressure by w

A gravity retaining wall shown in fig 1 is the one in which the earth pressure exrted by the

back fill is resisted by dead weight of wall, which is either made of masonry or of mass concrete . The

stress devlop in the wall is very low ,These walls are no proportioned that no tension is devloped any

where, and the resultant of forces remain withen the middle third of the base.

A retaining wall or retaining structure is used for maintaining the ground surfgaces at defrent

elevations on either side of it. Whenever embankments are involed in construction ,retaining wall are

usually necessary. In the construction of buildins having basements, retaining walls are mandatory.

Similsrly in bridge work, the wing walls and abutments etc. are designed as retaining walls , to resist

earth pressure along with superimposed loads. The material retained or supported by a retaining wall

is called backfill lying above the horizontal plane at the elevation of the top of a wall is called the

surcharge, and its inclination to horizontal is called the surcharge angle b

In the design of retaining walls or other retaining structures, it is necessary to compute the

lateral earth pressure exerted bythe retaining mass of soil. The equation of finding out the lateral earth

pressure against retaining wall is one of the oldest in Civil Engineering field. The plastic state of

strees, when the failure is imminent, was invetigated by Rankine in1860. A Lot of theoretical

experiment work has been done in this field and many theory and hypothesis heve benn proposed.

RETAINING WALL

Retaining walls may be classified according to their mode of resisting the earth pressure,and

according to their shape. Following are some of commen types of retaining walls (Fig)


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y of beending of varios components acting as cantilever s.A coomon form of cantilever retaining waal


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b Surcharge anglem

mm F

Hieght of cantilever wall from ground level = 3.00 m @ c/c

Unit weight of Earth = 18 KN/m3 m

Angle of repose = 30 Degree mm F

Safe Bearing capacity of soil = 100 KN/m3 @ c/cCoffiecent of friction = 0.5 m

Concrete M- 20 25000 N/m3 mm F

cbc 7 N/mm2 m 13.33 @ c/cSteel fe 415 N/mm2 st 230 N/mm2 m

Nominal cover = 30 mm

Surcharge angle b 16 Degree mFounadation depth = 1.00 m Toe

Stem thickness At footing 310 mm At top 200 mm

Heel width 900 mm Toe width 1200 mm mm F

Footing width 2100 Key 300 x 300 mm @ c/c

Reinforcement Summary

STEM :- mm F @ c/c m

Main

2.58 12 mm F@ 90 mm c/c 8 F mm F

1.94 12 mm F@ 180 mm c/c @ c/c

Top 12 mm F@ 360 mm c/c

Distribution 8 mm F@ 160 mm c/c

Tamprecture 8 mm F@ 300 mm c/c mm F

TOE :- @ c/c

Main 12 mm F@ 120 mm c/c

Distribution 8 mm F@ 170 mm c/c mm F

HEEL :- @ c/c

Main 10 mm F@ 210 mm c/c

Distribution 8 mm F@ 170 mm c/c

wt. of concrete

3730

DESIGN OF T SHAPED CANTILEVER RETAINING WALL

with sloping back fill 8

3000

300 3000

170

180

3730

100% Reinforcement upto m



8

DESIGN SUMMARY

12

30

12

20

8

160

12

360

90

12

360

90

12

180

210

12

8

10

12

Heel

200270

1000

Out side

300

Earth side

2.58

1.94

9001200 300

8

170

540

730

160

2100

120

3730

200

3002.58

1.94

8

4000

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Hieght of cantilever wall from ground level = 3.00 m

Unit weight of Earth g = 18 kN/m = N/m

Angle of repose = 30 Degree

Safe Bearing capacity of soil q0 = 100 kN/m3

Coffiecent of friction m = 0.5 = 25 N/mmConcrete = M 20

Steel fe = 415

Nominal cover = 30 mm

Surcharge angle = 16 Degree

Founadation depth = 1.00 m

1 Design Constants:-For HYSD Bars = 20

st = = 230 N/mm = #### N/mm

cbc = = 7 N/mm3

m = 13.33x

13.33 x 7 + 230j=1-k/3 = 1 - 0.289 / 3 = 0.904

R=1/2xc x j x k = 0.5 x 7 x 0.904 x 0.289 = 0.913

2 Diamension of base:-

sin b = 0.276 cos b = 0.96 tanb = 0.29Sin F = 0.5 Cos F = 0.87

cos b - cos2b -cos2f 0.961 - 0.92 - 0.75

cos b + cos2 b - cos2f 0.961 + 0.92 - 0.75

For surcharge wall, The ratio of length of slabe (DE) to base width b is given by eq.

q0

2.7 y H 2.7 x 18 x 4.00

The base width is given by Eq.

x

x x

( 1 - 0.49 )x( 1 + 1.46 )

x 4.00 x 0.38

( 1 - 0.49 )x 0.5

0.6 b = 0.60 x 4.00 = m

Hence Provided b = m

Width of toe slab = a x b = 0.49 x 2.40 = 1.17 m mLet the thickness of base be = H/12 = 4.00 / 12 = 0.33 or say = 0.30 m

= 2.40 - 0.30 - 1.20 = 0.90 m

3 Thickness of stem:-

= 4.00 - 0.30 = 3.70 m consider 1 m length of retaining wall

K x y x H12 0.38 x 18 x( 3.70 )

2

= 46.75 x 0.96 = 45 kN

46.75 Kn-m

=

Ka

Eq (1)

Hence the horizontal earth pressure is PH= P cos b

1.20

for design

purpose

=

The base width from the considration of sliding is given by Eq.

b

=

= 2.15

b

4.00

= 1 -

H

m

18000

100

=

= 0.289

2

0.7

=2

Hence width of heel slab

Heigth AB

Maximum Bending momentat B =

DESIGN OF T SHAPED CANTILEVER RETAINING WALL

Cocrete M

wt. of concrete

13.33 7m*c=k=

Provided toe slab =

m*c+sst

=

(1-a) m=

1

mb =

The wall will be unsafe against sliding. This will be made safe by providing a shear Key at base .

4.13

This width is excessive. Normal practice is to provide b between 0.4 to 0.6 H .

Taking maximum value of H = 2.40

=0.7HKa

2.40

0.49-

Ka cos b

(1- a) x (1+3 a)

= cos b 0.96 0.38

0.38 0.961

= =x

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H1 3.70

3 3

BM 55.50 x 10 6

Rxb 0.913 x 1000

= 250 mm and total thickness = 250 + 60 = 310 mm

Assuming that 12 mm F bar will be used. a nominal cover of = 60 - 6 = 54 mm= 200 mm at top so that effective depth of = 140 mm

45 x 1000

1000 x 250

4

= 2.40 - 1.20 - 0.31 = 0.89 m

= 3.70 + 0.89 x 0.29 = 3.96 m

= 3.96 + 0.30 = 4.26 m

0.38 x 18 x( 4.26 )2

2 2

PH = P cos b = 61.84 x 0.96 = 59.44 kN

PV = P sin b 61.84 x 0.28 = 17.04 kN

Full dimension wall is shown in fig 1a

Let W1 = weight of rectangular portion of stem

w2 = weight of triangular portion of stem

w3 = weight of base slab

w4 = weight of soil on heel slab.

The calculation are arrenged in Table

force(kN) Moment about toe (KN-m)

w1 1 x 0.20 x 3.70 x 25 =

w2 1/2 x 0.11 x 3.70 x 25 =

w3 1 x 2.40 x 0.30 x 25 =

w4 1 x 0.90 x 3.83 x 18 =

w5 = 2.40

Sw =Total resisting moment = kN-m ..(1)

4

3

mSw 0.5 xPH

- 87.7 = kN-m

\ Distance x of the point of application of resultant, from toe is

SM 128.18 b 2.40Sw 120.64 6 6b 2.40

2 2

SW 6 e 120.64 6x 0.14 67.55 < 100b b 2.40

SW 6 e 120.64 6x 0.14 32.99 < 100b b 2.40

67.55 - 32.99

=

x =

Pressure p at the junction of stem with toe slab is

Pressure p1 at

toe

=59.44

x

x 1 +

1 -

=

Height H2= H1+Ls tan b

Height H

P is acting on vertical face IG, at H/3 and hence Pv , will act the vertical line

40.90Pv 17.04

Earth pressure p=

Its horizontal and vertical component are

26.085

lever arm

kN

N/mm2 > tc even at mimum steel

Length of heel slab

tv = = 0.18

PH x = 45.00 x = 55.5

2.401.20 50.27

=2.40 kN -m

2

Hence safe

kN-m2

Pressure p' at the junction of stem with Heel slab is

p = 67.55 -

1.06 m ==

1.01 2

..(2)

Over turning moment Mo = 87.7 kN-m

= 61.84

x =61.8

total MR120.64

18

62.01

1.2

1.95

1.41

247

215.89

=

1.255

Detail

mm

18.5

5.09

Reduce the total thickness

=Effective depth required

B.M. at B =

128.18

=Ka x y x H

2

-

Pressure p1 at

Heel= 1 -

Eccenticity e = x

F.S. against Sliding

215.89

Pressure distribution net moment SM =

Over turning

\ F.S. against over turning

Hence not safe , To make safe against sliding will have to provide shear key

0.4

Keep d

=

=

x=

-

=

=

= 1 + =

= Hence safe0.14 0.4

2.40=

m 540 Hence safe

0.12 270 + 200100

P D2 3.14 x ( 8 )'2

41000 x 50

6 Design of heel slab :-Three force act on it

2 weight of heel slab 3 Down ward earth pressure 4 upward soil pressure

3.70 + 3.96

3.70 + 2 x 3.96 0.90

+ 3

= 0.90 x 0.27 x 1 x 25 = kN

Acting at 0.45 m from B.

Earth pressure intencity at b = Ka.y.H1 per unit inclined area, at b to horizontal,\ Earth pressure at B, on horizontal unitarea = Ka.y.H1.tan b

Vertical component of this, at B = Ka.y.H1 .tan b.sin b .(I)

.(II)

Hence total force due to vertical component of earth pressure is

0.38 x 18

= 1.86 kN m from B

63.00

= 0.455 m from B

=2

Ka.y(H1+H2)

2

(H1+2xH2)xb

(H1+H2)x3=

x

Total weight of heel slab

Acting at3.70

Total weight of

soil over Heel=

2xx0.90

1. down ward weight of soil

Using

12 mm bars

And at underE

x 0.90

282 mm2

2

Spacing =

8 mm F bars, Area

Distribution steel

sst x j x D

= x x

=

3.96

4mm say =

=

170

b1 tan b x sin b

=

120

120

50

mm c/c

mm2

x

18 kNKN say

6.08

The reinforcement has to be provided at bottom face .If alternate bars of stem reiforcerment are

= 62

using

are bent and continued in toe slab, area available (see step 7)628

1000

Reduce the total thickness to

=

Effective depth required =

BM x 106

Keep effective depth d

x from E

=

tv

896

= 0.23=

0.63

39.09mm

2

tc even at mimum steel

m

N/mm2 0.14

7 Reinforcement in the stem:-We had earliar assume the thickeness of heel slab as = 0.30 m

0.27 m only. Hence revised H1= 4.00 - 0.27 = 3.73 m

ka.y 0.38 x 18

2

S.F. x H1 47.52 x 3.73

= 250 mm and total thickness = 250 + 60 = 310 mm

= 200 mm or m at edge

x

230 x 0.904 x 250

P D2

3.14 x ( 12 )

'2

4

1000 x 113

113

90

Bend these bars into toe slab, to serve as reiforcement there. Sufficient devlopment length ia available.

+ 250 - 140

PH

3

)2=

2.0047.52

kN-m

x( 3.73

Hence Safe

S.F at B = pcos b = H12 = kN =

Spacing =

Using 12

The effective depth d' at section is =

=Actual AS provided =

Between A and B some of bars can be curtailed. Cosider a section at depth below the top of stem

1000 x

H140 x

90 mm c/c

(where h In meter)

1256 mm2

h

mm24

= 991137

=mm F bars, Area =

mm say =

106

= 1137 mm2

= 113

=

Keep effective depth d

Reduce the total thickness to 0.20

Ast = BMx100/sstxjxD=

=

0.14 N/mm2

%

While it has now been fixed as

= 0.15

steel provided tc N/mm2Permissible shear stress for

Shear stress tv

B.M. at B3

x 100

=

=shear force

=

If tc > tvhence safe

= 50

Nomber of Bars =

Hence Provided 10

% of steel provided =

Using 8 mm F bars, Area =

mm c/c

mm c/c170

210

mm2

mm

2

=

x2

= 288

288

=

= 174

4

= x 1000

mm saySpacing =

Hence provided these @

from a distance of

59.08

= 78.5

Spacing =

10Using

Distribution steel

=

=4

mm F bars, Area =

366= 214 mm say

Ast = 15.97 10

6

sst x j x DBM x 10

6

=

m from B

=

200

< tc even at mimum steel

29.67

mm2

mm2

39.41

0.41

59.08

tv

Acting at =54.59

Total upward soil pressure

S.F. at B

mm and D

This is much lessthan the B.M. on slab. However, we keep the same depth, as that toe slab,i.e.

mm, reducing it to

N/mm2

32.99

= = 0.14

= 366

N-mm2

x =

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250 - 140

H3

d'

h Ast' d'1/3

H1 st

where Ast' = reinforcement at depth h Ast = reinforcement at depth H1

d' = effective depthat depth h d = effective depthat depth H1

Ast' 1 h 1 d'1/3

Ast 2 H1 2 d

Subsituting d = 255 =( 140 + 29.5 x h ) we get

140 + 29.5 x h 1/3

x

140 x 29.5 x hx

h = 0.467 x ( 140 + 29.5 x h 1/3 ..(3)

h = 2.83 m 0.467 x ( 140 + 29.5 x h)

- h =

Howerver, the bars should be extented by a distance of 12 F = 12 x 12 = 144 mmOr d = 250 mm whichever is more beyond the point.

\ h = 2.83 - 0.25 = 2.58 m. Hence curtailed half bars at at height of2.58 m below the top . If we wish to curtailed half of the remaining bars so that remaining

Ast' 1Ast 4

h 1 x d' 140 + 29.5 x h 1/3

H1 4 d x

x 140 x 29.5 x h4 x

h = 0.371 x ( 140 + 29.5 x h 1/3 ..(4)

h = 2.19 m 0.371 x ( 140 + 29.5 x h) - h =

.Howerver, the bars should be extented by a distance of 12 F = 12 x 12 = 144 mmOr d = 250 mm whichever is more beyond the point.

\ h = 2.19 - 0.25 = 1.94 m. Hence stop half bars the remaining barsby 1.94 m below the top of the stem . Continue rest of the bars to the top of the stem

Check for shear:-

Shear force = 182

47.52 x 1000

1000 x 250

Ast/A = 1137 / 113 = 10.06 say = 11 No.

11 bars of mm F at Bottom11 x 113

1000 x 250

0.50 % = 0.3 (See Table 3.1)

here 0.30 > 0.19

Distribution and temprechure reinforcement:-

= 310 + 200

0.12

100

P D2 3.14 x ( 8 )'2

4

1000 x 50

= 8 = 300 mm c/cboth way in outer face

0.01

255

= H1

Hence

d' =( 140

=

Now As )1/3

=

Ast dor

x h

Ast H =(

140 x h )29.49++3.73

,,,'(1)

Hence Provided 12

255

x

h = 3.73

=\

2 255

mm and d'

x

h =

if Ast = than

H1

1/2 Ast

2

=

x

remaining reinforcement is one forth of that provided arB, we have

% of steel provided = x 100 = 0.50 %

Hence from .(2)=

Permissible shear stress for steel provided tc N/mm2

\ h

This can be solved bytrial and error, Noting that if the effective thickness of stem w=are constant,

x

0.00

\

h = 3.73

=4 255

= 0.38p = kNxkayH

2

23.73

2

=x

\ tc

If tc > tvhence safe Hence Safe

47.52

= 0.19 N/mm2 < (see table 3.1)

Average thickness of stem2

= 255

=tv

Nomber of Bars =

mm

\ Distribution reinforcement = x 1000 x 255 = 306

= = = 50.24 mm2

4

mm bars

mm c/c at the inner face ofwall,along its length

mm2

Using 8 mm F bars, Area

306= 164 mm say = 160

for tempreture reinforcement provide

\ spacing

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8 Design of shear key:-

300 x 310 Let Pp be the intensity of passive pressure devloped

in front of key this intencity Pp depend upon the soil pressure P in front of the key

Pp = KpP = 1/Ka= 1/ 0.38 = 2.64 x 50.27 = kN/m

Pp x a = x = 39.74 kN

18 `

2.00

or PH = 3.42 x( 4.53 )2x 0.961 = 67.22 kN

= 2.40 x 18 x 0.30 = 12.96 kN

\ W = 120.64 + 12.96 = kN Refer force calculation table=

mSw+Pp 0.5 x 133.60 + 39.74

= 300 mm. Keep width of key 310 mm (equal to stem width)

F where (45 + F/2) =2 shearing angle of passive resistance

\ a1 = 0.3 x ( 2.64 )1/2

a1 = 0.487 m = DE = 1.20 m

Hence satisfactory.

Now size of key = 300 x 310 mm

Actual force to be resisted by the key at F.S. 1.5 is = 1.5PH - mSW= 1.5 x 67.22 - 0.5 x

= kN

34.03 x 1000

300 x 1000

34.03 x 150 x 1000

1/6 x 1000 x( 300 )2

= N/mm2 Hence safe

Since concrete can take this much of tensile stress, no special reinforcement is necessary for the shear key

> 1.5 Hence safe

133.60

67.22= 1.58

Let u sprovide ashear key

0.30

The wall is in unsafe in sliding, and hence shear key will be provided below the stem as shown in fig.

132.46

\ total passive pressure Pp = 132.46

.(2)

Weight of the soil between bottom of the base and GJ

0.38 x x 4.53Sliding force at level GJ = cos bx

Hence equilibrium of wall, permitting F.S. 1.5 against sliding we have

PH=1.5 =

133.60

34.03

= 0.113 N/mm2

=

Actual length of the slab available

a kp45

Bending stress =

\ shear stress =

+

0.34

a tan F = a tana1 x

it should be noted that passive pressure taken into account above will be devloped only when length

a1 given below is avilable in front of key ;

However, provided minimum value of a

=

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10/16

b Surcharge angle b Surcharge angle b SurchA A A

m H1 m m H1= m m

toe heel

D E B C D E B C D E B

Toe Toe a1

m m

D1 e

p = pp

3.70

1.00

1.20

a b

W1

2.40

0.20

3.00 3.70

4.00H=

0.20

3.00

0.

0.20

3.00

2.40

W1

W2W2 0.90 0.00

2.40

P=

P=

0.30 0.30

b =b =

50.2

7

W1

2.40

1.20W2 0.90

67.5

5

50.2

7

54.5

9

32.9

9

67.5

5


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b Surcharge angleEarth side Face A Outer side face

mm F@ mm F `

@ c/c

m

mm F@

mm F@ c/c

mm F@

mm F@ c/c

mm FN.S.L. @

mm F mm F

@ c/c @

Heel Toe

Reinforcement Detail Reinforcemen

Foundation level

mm F mm F@ c/c @ c/c `

12

8

310

C/C

0.20

Outer sideEarth side Face

360

C/C

C/C

8

300 12

4.00H=

170

3.70

8

8

160

12

90

1.9

4

2.5

8

12

210 C/C

8

160 C/C

200

300

310

200

8

170120

180

9001200

1000

300

10


12/16


13/16


14/16

M-10 M-15 M-20 M-25 M-30 M-35 M-40

(N/mm2) Kg/m2

(N/mm2) Kg/m2

M 10 3.0 300 2.5 250

M 15 5.0 500 4.0 400

M 20 7.0 700 5.0 500

M 25 8.5 850 6.0 600

M 30 10.0 1000 8.0 800

M 35 11.5 1150 9.0 900

M 40 13.0 1300 10.0 1000

M 45 14.5 1450 11.0 1100M 50 16.0 1600 12.0 1200

M-10 M-15 M-20 M-25 M-30 M-35 M-40

Grade of concrete M-15 M-20 M-25 M-30 M-35 M-40

Modular Ratio 18.67 13.33 10.98 9.33 8.11 7.18

scbc N/mm2 5 7 8.5 10 11.5 13m scbc 93.33 93.33 93.33 93.33 93.33 93.33

kc 0.4 0.4 0.4 0.4 0.4 0.4

jc 0.867 0.867 0.867 0.867 0.867 0.867

Rc 0.867 1.214 1.474 1.734 1.994 2.254

Pc (%) 0.714 1 1.214 1.429 1.643 1.857

kc 0.329 0.329 0.329 0.329 0.329 0.329

jc 0.89 0.89 0.89 0.89 0.89 0.89

Rc 0.732 1.025 1.244 1.464 1.684 1.903

Pc (%) 0.433 0.606 0.736 0.866 0.997 1.127

kc 0.289 0.289 0.289 0.289 0.289 0.289

jc 0.904 0.904 0.904 0.904 0.904 0.904Rc 0.653 0.914 1.11 1.306 1.502 1.698

Pc (%) 0.314 0.44 0.534 0.628 0.722 0.816

kc 0.253 0.253 0.253 0.253 0.253 0.253

jc 0.916 0.916 0.916 0.914 0.916 0.916

Rc 0.579 0.811 0.985 1.159 1.332 1.506

Pc (%) 0.23 0.322 0.391 0.46 0.53 0.599

(d) sst =275

N/mm2

(Fe 500)

(c ) sst =

230N/mm2

(Fe 415)

(b) sst =190

N/mm2

Table 2.1. VALUES OF DESIGN CONSTANTS

(a) sst =140

N/mm2

(Fe 250)

Grade of concrete

Modular ratio m31

(31.11)

19

(18.67)

13

(13.33)

11

(10.98)

9

(9.33)

8

(8.11)

7

(7.18)

1.4 140

Table 1.18. MODULAR RATIO

1.1 110

1.2 120

1.3 130

0.8 80

0.9 90

1.0 100

2.8 3.2

-- --

0.6 60

4.4

Grade of

concrete

Permission stress in compression (N/mm2) Permissible stress in bond (Average) for

plain bars in tention (N/mm2)Bending acbc Direct (acc)

(N/mm2) in kg/m2

Table 1.15. PERMISSIBLE DIRECT TENSILE STRESS

Table 1.16.. Permissible stress in concrete (IS : 456-2000)

Grade of concrete

Tensile stress N/mm2 1.2 2.0 3.6 4.0


15/16

M-15 M-20 M-25 M-30 M-35 M-40 < %

0.15 % 0.18 0.18 0.19 0.20 0.20 0.20 %

0.25 % 0.22 0.22 0.23 0.23 0.23 0.23 %

0.50 % 0.29 0.30 0.31 0.31 0.31 0.32 %

0.75 % 0.34 0.35 0.36 0.37 0.37 0.38 %

1.00 % 0.37 0.39 0.40 0.41 0.42 0.42 %

1.25 % 0.40 0.42 0.44 0.45 0.45 0.46 %

1.50 % 0.42 0.45 0.46 0.48 0.49 0.49 %

1.75 % 0.44 0.47 0.49 0.50 0.52 0.52 %

2.00 % 0.44 0.49 0.51 0.53 0.54 0.55 %

2.25 % 0.44 0.51 0.53 0.55 0.56 0.57 %

2.50 % 0.44 0.51 0.55 0.57 0.58 0.60 %

2.75 % 0.44 0.51 0.56 0.58 0.60 0.62

3.00 and above % 0.44 0.51 0.57 0.6 0.62 0.63

300 or more 275 250 225 200 175 50 or less

1.00 1.05 1.10 1.15 1.20 1.25 1.30

M-15 M-20 M-25 M-30 M-35 M-40

1.6 1.8 1.9 2.2 2.3 2.5

Grade of concrete M-10 M-15 M-20 M-25 M-30 M-35 M-40 M-45 M-50

tbd (N / mm2) -- 0.6 0.8 0.9 1 1.1 1.2 1.3 1.4

tbd (N / mm2) tbd (N / mm2)

M 15

M 20

M 25

M 30

M 35

M 40M 45

M 50

100As

bd

Table 3.4. Permissible Bond stress Table bd in concrete (IS : 456-2000)

1.4 25 2.24 26

1.3 27 2.081.2 29 1.92 30

28

331.1 32 1.76

40

1 35 1.6 36

0.9 39 1.44

60

0.8 44 1.28 45

0.6 58 0.96

Grade of concrete

tc.max

Table 3.5. Development Length in tension

Grade of

concrete

Plain M.S. Bars H.Y.S.D. Bars

kd = LdF kd = LdF

Table 3.3. Maximum shear stress tc.max in concrete (IS : 456-2000)

Table 3.2. Facor k

Over all depth of slab

k

Table 3.1. Permissible shear stress Table c in concrete (IS : 456-2000)

Permissible shear stress in concrete tc N/mm2


16/16

Degree sin cos tan

10 0.174 0.985 0.176

15 0.259 0.966 0.268

16 0.276 0.961 0.287

17 0.292 0.956 0.306

18 0.309 0.951 0.325

19 0.326 0.946 0.344

20 0.342 0.940 0.364

21 0.358 0.934 0.384

22 0.375 0.927 0.404

23 0.391 0.921 0.424

24 0.407 0.924 0.445

25 0.422 0.906 0.466

30 0.500 0.866 0.577

35 0.573 0.819 0.700

40 0.643 0.766 0.839

45 0.707 0.707 1.000

50 0.766 0.643 1.192

55 0.819 0.574 1.428

60 0.866 0.500 1.732

65 0.906 0.423 2.145

Value of angle

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